JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Definite Integrals

The value of

\sum_{n = 1}^{100} \int_{n - 1}^{n} e^{x - [x]} d x

, where

[x]

is the greatest integer

\leq x

, is:

Answer: (D)

Solution

For

n - 1 \leq x < n

, we have

[x] = n - 1

, so

x - [x] = x - (n - 1)

. Thus,

\int_{n - 1}^{n} e^{x - [x]} d x = \int_{n - 1}^{n} e^{x - (n - 1)} d x

. Let

u = x - (n - 1)

, then

d u = d x

, and when

x = n - 1

u = 0

; when

x = n

u = 1

. So the integral becomes

\int_{0}^{1} e^{u} d u = [e^{u}]_{0}^{1} = e - 1

. Since this is the same for each

n

from 1 to 100, the sum is

100 (e - 1)

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option

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